Question

If \[y = \cos^{- 1} \left( 2x \right) + 2 \cos^{- 1} \sqrt{1 - 4 x^2}, - \frac{1}{2} < x < 0, \text{ find } \frac{dy}{dx} \] ?

Answer 1

\[\text{ Here }, y =  \cos^{- 1} \left( 2x \right) + 2   \cos^{- 1} \sqrt{1 - 4 x^2}\]

\[\text{ Put }2x = \cos\theta\]

\[ \therefore y =  \cos^{- 1} \left( \cos  \theta \right) + 2   \cos^{- 1} \sqrt{1 - \cos^2 \theta}\]

\[ \Rightarrow y =  \cos^{- 1} \left( \cos  \theta \right) + 2   \cos^{- 1} \left( \sin\theta \right)\]

\[ \Rightarrow y =  \cos^{- 1} \left( \cos  \theta \right) + 2   \cos^{- 1} \left[ \cos\left( \frac{\pi}{2} - \theta \right) \right] ........\left( 1 \right)\]

\[\text{Now, }-\frac{1}{2} < x < 0\]

\[ \Rightarrow - 1 < 2x < 0\]

\[ \Rightarrow - 1 < \cos\theta < 0\]

\[ \Rightarrow \frac{\pi}{2} < \theta < \pi\]

And

\[ \Rightarrow - \frac{\pi}{2} >  - \theta >  - \pi\]

\[ \Rightarrow \left( \frac{\pi}{2} - \frac{\pi}{2} \right) > \left( \frac{\pi}{2} - \theta \right) > \left( \frac{\pi}{2} - \pi \right)\]

\[ \Rightarrow 0 > \left( \frac{\pi}{2} - \theta \right) >  - \frac{\pi}{2}\] 

\[ \Rightarrow - \frac{\pi}{2} < \left( \frac{\pi}{2} - \theta \right) < 0\] 

\[\text{ So,   from  equation }  \left( 1 \right), \] 

\[y = \theta + 2\left[ - \left( \frac{\pi}{2} - \theta \right) \right]...........[\text{ Since }, \cos^{- 1} \cos\left( \theta \right) = \theta,\text{ if } \theta \in \left[ 0, \pi \right],  \cos^{- 1} \cos\left( \theta \right) = - \theta, \text{ if } \theta \in \left[ - \pi, 0 \right]]\] 

\[y = \theta - 2 \times \frac{\pi}{2} + 2\theta\] 

\[y =  - \pi + 3\theta\] 

\[y =  - \pi + 3 \cos^{- 1} \left( 2x \right)..........\left[ \text{ Since }, 2x = cos\theta \right]\]

Differentiate it with respect to x using chain rule,

\[\frac{d y}{d x} = 0 + 3\left( \frac{- 1}{\sqrt{1 - \left( 2x \right)^2}} \right)\frac{d}{dx}\left( 2x \right)\]

\[ \Rightarrow \frac{d y}{d x} = \frac{- 3}{\sqrt{1 - 4 x^2}} \times 2\]

\[ \therefore \frac{d y}{d x} = - \frac{6}{\sqrt{1 - 4 x^2}}\]